The ratio and root tests
Factorials and powers make terms that are hard to compare with anything, yet the ratio of one term to the next is often simple. When does that ratio, or the nth root of a term, settle whether a series converges, and when does it say nothing?
Where you are 6 in Convergence tests
Why this exists
Chipo runs a banquet hall with one long table that has a garden side and a wall side. At a banquet of 2n guests, each guest picks a side independently, the garden side with probability p. There are \binom{2n}{n} ways to choose which n guests sit by the garden (The binomial theorem), so the table splits evenly, n on each side, with probability (The binomial probability formula)
Chipo hosts one banquet of each size n = 1, 2, 3, \ldots and asks how many evenly split banquets to expect. Expected counts add, so the answer is the sum of these probabilities: an infinite series.
For \displaystyle p = \frac15 the terms are \displaystyle \frac{(2n)!}{(n!)^2}\left(\frac{4}{25}\right)^n, a factorial count growing against a power shrinking. To use comparison one must guess a benchmark, and here the right one is not obvious. But the ratio of one term to the next is simple, and it settles toward \displaystyle \frac{16}{25}. A series whose terms shrink by a factor near \displaystyle \frac{16}{25} at each step should behave like a geometric series with that ratio. The ratio test makes this precise, and the root test does the same with \displaystyle \sqrt[n]{\lvert a_n \rvert}, which suits terms raised to a power that grows with n. Both compare \lvert a_n \rvert with a geometric series, so what they prove is absolute convergence.
Definition
Day 1: the ratio test
Let \sum a_n be a series with a_n \ne 0 from some index on, and suppose the limit
exists as a number or is \infty. In words: compare each term’s size with the size of the term before it, and see what that factor approaches. The ratio test says:
- if L < 1, the series converges absolutely;
- if L > 1 or L = \infty, the series diverges;
- if L = 1, the test gives no conclusion.
Day 2: the root test
For a series \sum a_n, suppose \displaystyle L = \lim_{n \to \infty} \sqrt[n]{\lvert a_n \rvert} exists as a number or is \infty. In words: find the geometric rate r for which \lvert a_n \rvert is roughly r^n. The root test draws the same three conclusions: absolute convergence if L < 1, divergence if L > 1 or L = \infty, and nothing if L = 1.
Intuition
Day 1: ratios that settle
A geometric series has the same ratio at every step. The ratio test handles series whose ratio settles only in the long run. If the ratios approach L < 1, then far enough out every ratio is below a fixed r < 1, and from there on the terms are trapped under a geometric series with ratio r. The early terms, however large, are finitely many and cannot affect convergence. If the ratios approach L > 1, the terms eventually grow, and a series whose terms do not tend to 0 diverges.
Figure 1 shows why L = 1 is different. For \displaystyle p = \frac15, Chipo’s ratios climb toward \displaystyle \frac{16}{25}. For a fair choice, \displaystyle p = \frac12, the ratios also stay below 1, so every term is smaller than the one before, but they climb toward 1. The factor by which the terms shrink fades away, and no single geometric series stays above them. Example 3 shows that this second series diverges.
Day 2: the average rate
Write \rho_k = \lvert a_{k+1}/a_k \rvert for the ratio at step k. Then \lvert a_n \rvert = \lvert a_1 \rvert\,\rho_1 \rho_2 \cdots \rho_{n-1}, so
The first factor tends to 1, and the second behaves like the geometric mean of the ratios. So the ratio test reads the rate of shrinking at each step, and the root test reads the average rate. When the step rates settle, the average settles at the same value. When the step rates keep jumping, the average can still settle (Example 7), so the root test reaches some series the ratio test cannot. In practice the ratio test suits factorials, which cancel in a ratio, and the root test suits an exponent that grows with n.
Results
The ratio test. If a_n \ne 0 from some index on and \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \to L, then \sum a_n converges absolutely when L < 1 and diverges when L > 1 or L = \infty.
The root test. If \displaystyle \sqrt[n]{\lvert a_n \rvert} \to L, then \sum a_n converges absolutely when L < 1 and diverges when L > 1 or L = \infty.
Eventually below a fixed rate. If there are r < 1 and N with \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \le r for every n \ge N, or with \displaystyle \sqrt[n]{\lvert a_n \rvert} \le r for every n \ge N, then \sum a_n converges absolutely. No limit is needed.
Tail bounds. If \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \le r < 1 for every n \ge N+1, then
If \displaystyle \sqrt[n]{\lvert a_n \rvert} \le r < 1 for every n \ge N+1, then \sum_{n=N+1}^{\infty} \lvert a_n \rvert \le \dfrac{r^{N+1}}{1-r}. Either sum bounds \lvert S - S_N \rvert.
A ratio limit is also a root limit (stated without proof). If a_n \ne 0 and \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \to L, then \displaystyle \sqrt[n]{\lvert a_n \rvert} \to L. The converse is false (Example 7).
L = 1 decides nothing. For every p series \displaystyle \sum \frac{1}{n^p} both limits equal 1, yet the series converges for p > 1 and diverges for p \le 1.
Why it is true
Ratio test, L < 1. Choose r with L < r < 1, for example \displaystyle r = \frac{L+1}{2}. Since the ratios tend to L, there is an N with \lvert a_{n+1} \rvert \le r\lvert a_n \rvert for every n \ge N. Applying this k times,
The series \sum_k \lvert a_N \rvert r^k is geometric with ratio r < 1, so it converges, and by direct comparison so does \sum_{n \ge N} \lvert a_n \rvert. Adding the finitely many earlier terms does not change convergence, so \sum a_n converges absolutely, and therefore converges. Starting the same chain at N+1 gives \sum_{n > N} \lvert a_n \rvert \le \lvert a_{N+1} \rvert (1 + r + r^2 + \cdots), the tail bound. The argument used only that the ratios are eventually at most r, which is the “eventually below a fixed rate” result.
Ratio test, L > 1 or L = \infty. Then \lvert a_{n+1} \rvert > \lvert a_n \rvert for every n from some N on, so \lvert a_n \rvert \ge \lvert a_N \rvert > 0 for n \ge N. The terms do not tend to 0, and the series diverges by the The nth term test for divergence.
Root test. If L < 1, choose r as before; from some N on, \displaystyle \sqrt[n]{\lvert a_n \rvert} \le r, that is \lvert a_n \rvert \le r^n, and direct comparison with \sum r^n gives absolute convergence, with tail \displaystyle \sum_{n > N} r^n = \frac{r^{N+1}}{1-r}. If L > 1, then \displaystyle \sqrt[n]{\lvert a_n \rvert} > 1, so \lvert a_n \rvert > 1, from some index on, and the series diverges by the nth term test.
Why L = 1 is inconclusive. For \displaystyle a_n = \frac{1}{n^p} the ratio is \displaystyle \left(\frac{n}{n+1}\right)^p \to 1. For the root, \displaystyle \ln \sqrt[n]{n} = \frac{\ln n}{n} \to 0 (L’Hôpital’s rule on \displaystyle \frac{\ln x}{x} gives the limit of \displaystyle \frac{1/x}{1}), so \displaystyle \sqrt[n]{n} \to 1 and \displaystyle \sqrt[n]{1/n^p} = \bigl(\sqrt[n]{n}\bigr)^{-p} \to 1. By the The integral test and p series these series converge for p > 1 and diverge for p \le 1, so a limit of 1 is reached by both kinds.
Worked examples
Day 1: Chipo’s garden-side banquets
Show that \sum_{n=1}^{\infty} \dfrac{(2n)!}{(n!)^2}\left(\dfrac{4}{25}\right)^n converges, and use the tail bound with S_5 to locate its sum.
SolutionSince (2n+2)! = (2n+2)(2n+1)\,(2n)! and (n+1)! = (n+1)\,n!, the factorials cancel in the ratio:
Dividing top and bottom by n, the ratio tends to \displaystyle \frac{8 \cdot 2}{25} = \frac{16}{25} < 1, so the series converges. Its terms are positive, so convergence and absolute convergence coincide.
For the tail, write the ratio as \displaystyle \frac{8}{25}\left(2 - \frac{1}{n+1}\right), which is below \displaystyle \frac{16}{25} for every n. So the tail bound holds with \displaystyle r = \frac{16}{25} from any N. The first terms are 0.32, 0.1536, 0.08192, 0.0458752 and 0.0264241, so S_5 \approx 0.62782, and \displaystyle a_6 = 924\left(\frac{4}{25}\right)^6 \approx 0.0155021. Then
so 0.6278 < S < 0.6709: Chipo expects fewer than one evenly split banquet over all sizes. The bounds are all this lesson needs. For comparison only: a later lesson, The binomial series, proves that \displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}} for \displaystyle \lvert x \rvert < \frac14. Quoting that result here without proof, the value \displaystyle x = \frac{4}{25} gives \displaystyle \frac{1}{\sqrt{9/25}} = \frac53, and removing the n = 0 term, 1, leaves \displaystyle S = \frac23 \approx 0.6667, inside the bounds.
Day 1: a long climb before the fall
Decide whether \sum_{n=1}^{\infty} \dfrac{n^{10}}{1.1^n} converges, and find its largest term.
SolutionThe ratio is
which tends to \displaystyle \frac{1}{1.1} = \frac{10}{11} < 1. The series converges.
The terms do not start shrinking at once. The ratio exceeds 1 exactly when (1 + 1/n)^{10} > 1.1, that is when \displaystyle 1 + \frac1n > 1.1^{1/10} \approx 1.009577, that is when \displaystyle n < \frac{1}{0.009577} \approx 104.4. So a_{n+1} > a_n for n \le 104 and a_{n+1} < a_n for n \ge 105, and the largest term is
The terms stay at least 1 until n = 685 (Figure 2), and the sum is about 6.155 \times 10^{17}. A table of the first hundred terms would suggest divergence. The ratio test looks only at the far end, where (beyond n = 200, for instance) every ratio is below 0.96, and that is where convergence is decided.
Day 1: a fair choice, and a limit of 1
With \displaystyle p = \frac12, Chipo’s series is \sum_{n=1}^{\infty} \binom{2n}{n}\dfrac{1}{4^n}. Show that the ratio test gives no conclusion, and decide whether the series converges.
SolutionThe factorials cancel as in Example 1, now with \displaystyle \frac14 in place of \displaystyle \frac{4}{25}:
The test says nothing. Every ratio is below 1, so the terms decrease, but that alone proves nothing: the harmonic series also has decreasing terms.
Compare the ratios with those of \displaystyle b_n = \frac1n, which are \displaystyle \frac{b_{n+1}}{b_n} = \frac{n}{n+1} = \frac{2n}{2n+2}, smaller than \displaystyle \frac{2n+1}{2n+2}. Hence
so \displaystyle \frac{a_n}{b_n} increases with n and is at least \displaystyle \frac{a_1}{b_1} = \frac12. That is, \displaystyle a_n \ge \frac{1}{2n}. The harmonic series diverges, so by direct comparison this series diverges: with a fair choice Chipo expects infinitely many evenly split banquets over all sizes. Two series with ratio limit 1, \displaystyle \sum \frac1n with ratios \displaystyle \frac{n}{n+1} and \displaystyle \sum \frac{1}{n^2} with ratios \displaystyle \left(\frac{n}{n+1}\right)^2, show the same limit for a divergent and a convergent series.
Day 1: a parameter, and the case the test leaves open
For which c > 0 does \sum_{n=1}^{\infty} (-1)^n \dfrac{c^n}{\binom{3n}{n}} converge?
SolutionHere \displaystyle \binom{3n+3}{n+1} = \frac{(3n+3)!}{(n+1)!\,(2n+2)!}, and dividing by \displaystyle \binom{3n}{n} = \frac{(3n)!}{n!\,(2n)!} leaves three factors on top and three below:
The absolute ratio of the series is c divided by this, so it tends to \displaystyle \frac{4c}{27}. The series converges absolutely for \displaystyle c < \frac{27}{4} and diverges for \displaystyle c > \frac{27}{4}.
At \displaystyle c = \frac{27}{4} the limit is 1 and another argument is needed. By the binomial theorem, \displaystyle 1 = \left(\frac13 + \frac23\right)^{3n} is a sum of positive terms, one of which is \displaystyle \binom{3n}{n}\left(\frac13\right)^n\left(\frac23\right)^{2n} = \binom{3n}{n}\frac{4^n}{27^n}. So \displaystyle \binom{3n}{n} \le \left(\frac{27}{4}\right)^n, and
The terms do not tend to 0, so the series diverges. It converges exactly when \displaystyle 0 < c < \frac{27}{4}, and then absolutely; for no c does it converge conditionally.
Day 2: an exponent of n squared
Decide whether \sum_{n=1}^{\infty} \left(\dfrac{n}{n+3}\right)^{n^2} converges.
SolutionThe exponent n^2 invites the root test. Since the terms are positive,
Its logarithm is \displaystyle -n\ln\left(1 + \frac3n\right) = -\frac{\ln(1 + 3t)}{t} with \displaystyle t = \frac1n \to 0^+, a \displaystyle \frac00 form. L’Hôpital’s rule gives the limit of \displaystyle -\frac{3/(1+3t)}{1}, which is -3. So \displaystyle \sqrt[n]{a_n} \to e^{-3} \approx 0.0498 < 1, and the series converges. The terms fall fast: 0.25, 0.0256, 0.00195, 0.000129. The ratio test would need the limit of \displaystyle \left(\frac{n+1}{n+4}\right)^{(n+1)^2} divided by \displaystyle \left(\frac{n}{n+3}\right)^{n^2}, a much harder computation.
Day 2: a root that tends to 0, with a tail bound
Show that \sum_{n=1}^{\infty} \dfrac{2^n}{n^{n/2}} converges, and find its sum to three decimal places.
SolutionThe root is \displaystyle \sqrt[n]{a_n} = \frac{2}{n^{1/2}} = \frac{2}{\sqrt n} \to 0, so the series converges.
For the sum, \displaystyle \frac{2}{\sqrt n} \le \frac12 exactly when \displaystyle \sqrt n \ge 4, that is for every n \ge 16. The root tail bound with N = 15 and \displaystyle r = \frac12 gives
The terms begin 2, 2, 1.5396, 1, 0.5724, 0.2963, \ldots and fall quickly, with S_{15} \approx 7.65401. So 7.65401 < S < 7.65405, and S \approx 7.654 to three decimal places.
Day 2: when the ratio jumps but the root settles
Decide whether \sum_{n=1}^{\infty} \dfrac{3^{(-1)^n}}{2^n} = \dfrac16 + \dfrac34 + \dfrac{1}{24} + \dfrac{3}{16} + \cdots converges, and find its sum.
SolutionFor odd n the term is \displaystyle \frac{1}{3 \cdot 2^n}, for even n it is \displaystyle \frac{3}{2^n}. From an odd n to the next index the ratio is \displaystyle \frac{3}{2^{n+1}}\cdot 3 \cdot 2^n = \frac92; from an even n it is \displaystyle \frac{1}{3 \cdot 2^{n+1}}\cdot\frac{2^n}{3} = \frac{1}{18}. The ratios alternate between \displaystyle \frac92 and \displaystyle \frac{1}{18}, so they have no limit, and they exceed 1 infinitely often, so not even the “eventually below a fixed rate” form applies.
The root is \displaystyle \sqrt[n]{a_n} = \frac{3^{\pm 1/n}}{2}, and 3^{1/n} \to 1, so \displaystyle \sqrt[n]{a_n} \to \frac12 < 1. The series converges, absolutely. The average-rate picture explains it: the geometric mean of the two step rates is \displaystyle \sqrt{\frac92 \cdot \frac{1}{18}} = \frac12.
The terms are positive, so they may be summed in two groups, each geometric. Even n = 2k: \displaystyle \sum_{k=1}^{\infty} \frac{3}{4^k} = 3 \cdot \frac{1/4}{1 - 1/4} = 1. Odd n = 2k+1: \displaystyle \sum_{k=0}^{\infty} \frac{1}{3 \cdot 2^{2k+1}} = \frac{1/6}{1 - 1/4} = \frac29. The sum is \displaystyle 1 + \frac29 = \frac{11}{9}.
Boundaries
- L = 1 gives no information. It happens for every p series and, by Exercise 11, for every series whose terms are a rational function of n. Such series need comparison, the integral test, or the nth term test.
- A fixed rate below 1 is needed. Ratios that are each below 1 do not suffice: the harmonic series has ratios \displaystyle \frac{n}{n+1} < 1 and diverges. What the proof needs is one r < 1 that eventually bounds every ratio (or every root).
- The ratio test needs nonzero terms, at least from some index on, since \displaystyle \frac{a_{n+1}}{a_n} must be defined.
- Both tests detect only absolute convergence. When L < 1 the convergence is absolute, and when L > 1 the terms do not even tend to 0. So a conditionally convergent series always gives L = 1 or no limit, and must be handled by other tests, such as the alternating series test.
- Convergence says nothing about the early terms. Example 2 converges with terms above 10^{15}. A tail bound needs a rate r that holds from the chosen N on, not just in the limit.
Where this goes next
A power series \sum c_n (x-a)^n is a series whose terms carry a variable, and the ratio test, applied with x fixed, shows it converges absolutely for x near a and diverges far away. At the boundary between the two the limit is exactly 1, the case these tests leave open, and Power series and the interval of convergence settles those points with the other tests of this unit. The same ratios govern how fast a Taylor and Maclaurin series converges. Choosing a convergence test collects all the tests and reads the shape of the terms to pick one: factorials and nth powers point here.
Exercises
Day 1: the ratio test
Use the ratio test to decide whether each series converges absolutely or diverges.
- \sum_{n=1}^{\infty} \dfrac{5^n}{n^4\, 3^n}
- \sum_{n=1}^{\infty} \dfrac{6^n}{(n+2)!}
- \sum_{n=1}^{\infty} \dfrac{(n+1)!}{4^n\, n^2}
- \sum_{n=1}^{\infty} (-1)^n\dfrac{(2n)!}{(n!)^2\, 5^n}
- \sum_{n=1}^{\infty} \dfrac{(-4)^n\, n}{(n+1)!}
Day 1: when the limit is 1
- Show the ratio test is inconclusive for \sum_{n=1}^{\infty} \dfrac{n^2+1}{n^4+3}, then decide convergence another way.
- Show the ratio test is inconclusive for \displaystyle \sum_{n=1}^{\infty} \dfrac{n}{\sqrt{n^3+5}}, then decide convergence another way.
- Tatenda seats 3n guests at three tables, each guest picking a table at random, each with probability \displaystyle \frac13. The chance of an even split, n at each table, is \dfrac{(3n)!}{(n!)^3\, 27^n}. Show the ratio test says nothing about the sum of these chances over n \ge 1, and decide convergence by comparing ratios with those of \displaystyle \frac1n, as in Example 3.
Day 1: long climbs and tails
- For \sum_{n=1}^{\infty} \dfrac{n^{20}}{2^n}, decide convergence and find which term is largest.
- For \sum_{n=1}^{\infty} \dfrac{5^n}{(n+1)!}, compute S_{10} and use the ratio tail bound with \displaystyle r = \frac{5}{13} to bound S - S_{10} and locate the sum.
Day 1: reasoning
- Show that the ratio test gives L = 1 for every series \displaystyle \sum \frac{P(n)}{Q(n)} in which P and Q are nonzero polynomials.
- Chipo says: “If \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert < 1 for every n, the series converges.” Give a counterexample other than the harmonic series.
- Suppose a_n > 0 and \displaystyle \frac{a_{n+1}}{a_n} \to L < 1. Prove that \sum n\,a_n converges.
Day 2: the root test
Use the root test to decide whether each series converges absolutely or diverges, or say why it is inconclusive and settle the question another way.
- \sum_{n=2}^{\infty} \dfrac{5^n}{(\ln n)^n} (also find the last n with a_n > 1)
- \sum_{n=1}^{\infty} \dfrac{(-1)^n}{3^n}\left(1 + \dfrac1n\right)^{n^2}
- \sum_{n=1}^{\infty} \dfrac{n^n}{2^{n^2}}
- \sum_{n=1}^{\infty} \dfrac{(n+1)^{2n}}{(3n^2+1)^n}
- \sum_{n=1}^{\infty} \left(\dfrac{2\arctan n}{3}\right)^n
- \sum_{n=1}^{\infty} \left(\dfrac{n^2}{n^2+1}\right)^n
Day 2: tails and the two tests
- Nyasha sums \sum_{n=1}^{\infty} \left(\dfrac{4}{n+2}\right)^n. Compute S_6 and use the root tail bound to locate the sum.
- Show that the ratio test fails for \sum_{n=1}^{\infty} \dfrac{n^{(-1)^n}}{3^n} (the ratios have no limit and exceed 1 infinitely often), and that the root test settles it.
Day 2: reasoning
- Prove the root tail bound: if \displaystyle \sqrt[n]{\lvert a_n \rvert} \le r < 1 for every n \ge N+1, then \displaystyle \sum_{n=N+1}^{\infty} \lvert a_n \rvert \le \frac{r^{N+1}}{1-r}.
- Nyasha says that when the ratio test gives L = 1, there is no point trying the root test. Explain why she is right whenever the ratio limit exists.
Answers
- Diverges. \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac53\left(\frac{n}{n+1}\right)^4 \to \frac53 > 1.
- Converges absolutely. The ratio is \displaystyle \frac{6}{n+3} \to 0.
- Diverges. The ratio is \displaystyle \frac{n+2}{4}\cdot\frac{n^2}{(n+1)^2} \to \infty.
- Converges absolutely. As in Example 1, \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{2(2n+1)}{5(n+1)} \to \frac45.
- Converges absolutely. \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{4(n+1)}{n(n+2)} \to 0.
- Converges. The ratio is \displaystyle \frac{(n+1)^2+1}{n^2+1} times \displaystyle \frac{n^4+3}{(n+1)^4+3}, which tends to 1. By limit comparison with \displaystyle \frac{1}{n^2}: \displaystyle \frac{n^4+n^2}{n^4+3} \to 1, and \displaystyle \sum \frac{1}{n^2} converges.
- Diverges. The ratio \displaystyle \frac{n+1}{n}\sqrt{\frac{n^3+5}{(n+1)^3+5}} \to 1. By limit comparison with \displaystyle \frac{1}{\sqrt n}: \displaystyle \frac{n^{3/2}}{\sqrt{n^3+5}} \to 1, and \displaystyle \sum \frac{1}{\sqrt n} diverges (\displaystyle p = \frac12).
- Diverges. \displaystyle \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)}{27(n+1)^3}, which simplifies to \displaystyle \frac{(3n+2)(3n+1)}{9(n+1)^2} \to 1. Also \displaystyle \frac{(3n+2)(3n+1)}{9(n+1)^2} \ge \frac{n}{n+1}, since this says 9n^2 + 9n + 2 \ge 9n^2 + 9n. So \displaystyle \frac{a_n}{1/n} never decreases, and \displaystyle a_n \ge \frac{a_1}{n} = \frac{2}{9n}; the harmonic series diverges.
- Converges; the largest term is a_{29} \approx 3.30 \times 10^{20}. The ratio \displaystyle \frac{(1+1/n)^{20}}{2} \to \frac12. It exceeds 1 exactly when \displaystyle 1 + \frac1n > 2^{1/20} \approx 1.03526, that is n < 28.4: the terms rise through a_{29} and fall after.
- S_{10} \approx 28.3208, S - S_{10} \le 0.1656, so 28.3207 < S < 28.4865. The ratio is \displaystyle \frac{5}{n+2}, at most \displaystyle \frac{5}{13} for n \ge 11, so the tail is at most \displaystyle \frac{a_{11}}{1 - 5/13} = \frac{13}{8}a_{11}, with \displaystyle a_{11} = \frac{5^{11}}{12!} \approx 0.101937. (For comparison only: the series for e^x, proved in The common Maclaurin series, gives the exact sum \displaystyle S = \frac{e^5 - 6}{5} \approx 28.4826, inside these bounds.)
- \displaystyle \frac{a_{n+1}}{a_n} = \frac{P(n+1)}{P(n)}\cdot\frac{Q(n)}{Q(n+1)}. If P has degree d and leading coefficient p, dividing P(n+1) and P(n) by n^d sends both to p, so \displaystyle \frac{P(n+1)}{P(n)} \to 1; the same holds for Q. So L = 1.
- \displaystyle \sum \frac{1}{\sqrt n}: its ratios \displaystyle \sqrt{\frac{n}{n+1}} are all below 1, but it is a divergent p series (\displaystyle p = \frac12). The ratios tend to 1, so no fixed r < 1 bounds them.
- \displaystyle \frac{(n+1)a_{n+1}}{n a_n} = \frac{n+1}{n}\cdot\frac{a_{n+1}}{a_n}, which tends to 1 \cdot L = L < 1, so \sum n a_n converges by the ratio test.
- Converges; the last term above 1 is a_{148}. \displaystyle \sqrt[n]{a_n} = \frac{5}{\ln n} \to 0. And a_n > 1 exactly when \ln n < 5, that is n < e^5 \approx 148.4.
- Converges absolutely. \displaystyle \sqrt[n]{\lvert a_n \rvert} = \frac{(1 + 1/n)^n}{3} \to \frac{e}{3} \approx 0.906 < 1.
- Converges. \displaystyle \sqrt[n]{a_n} = \frac{n}{2^n} \to 0.
- Converges. \displaystyle \sqrt[n]{a_n} = \frac{(n+1)^2}{3n^2+1} \to \frac13.
- Diverges. \displaystyle \sqrt[n]{a_n} = \frac{2\arctan n}{3} \to \frac{2}{3}\cdot\frac{\pi}{2} = \frac{\pi}{3} \approx 1.047 > 1.
- Diverges. \displaystyle \sqrt[n]{a_n} = \frac{n^2}{n^2+1} \to 1, so the root test is inconclusive. But \displaystyle \ln a_n = -n\ln\left(1 + \frac{1}{n^2}\right) lies between \displaystyle -\frac1n and 0, since 0 < \ln(1+t) \le t for t > 0; so \ln a_n \to 0, a_n \to 1, and the nth term test gives divergence.
- S_6 \approx 3.1194 and 3.1194 < S < 3.1256. The root is \displaystyle \frac{4}{n+2}, at most \displaystyle \frac49 for n \ge 7, so \displaystyle S - S_6 \le \frac{(4/9)^7}{1 - 4/9}, which is \displaystyle \frac95\left(\frac49\right)^7 \approx 0.00617. (The true sum is about 3.1236.)
- Converges absolutely. For odd n, \displaystyle a_n = \frac{1}{n\,3^n} and \displaystyle a_{n+1} = \frac{n+1}{3^{n+1}}, so the ratio is \displaystyle \frac{n(n+1)}{3}, which is 4 at n = 3 and grows; for even n the ratio is \displaystyle \frac{1}{3n(n+1)} \to 0. The root is \displaystyle \frac{n^{\pm 1/n}}{3} \to \frac13, since \displaystyle \sqrt[n]{n} \to 1.
- For n \ge N+1, \lvert a_n \rvert \le r^n. So \sum_{n=N+1}^{\infty} \lvert a_n \rvert \le \sum_{n=N+1}^{\infty} r^n, a geometric series with first term r^{N+1} and ratio r, whose sum is \displaystyle \frac{r^{N+1}}{1-r}.
- If \displaystyle \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \to 1, then by the result that a ratio limit is also a root limit, \displaystyle \sqrt[n]{\lvert a_n \rvert} \to 1 as well, so the root test is inconclusive too. The root test can help only when the ratios have no limit, as in Example 7.